Fermat's version of the solution to the problem of points was to create a grid of possibilities reaching fully out to the point beyond which no doubt could exist as to who the winner would be. This grid of possibilities included parts of the tree which, on one view, would be utterly irrelevant to the game in hand, and on another view, incorrectly modelled the set of possibilities embedded in the game.
Pascal's solution, by way of contrast, was a ragged tree of possibilities stretching out along each branch only as far as was needed to resolve the state of the game in question, and no further.
Pascal additionally made the mistake, in interpreting Fermat's solution, of ignoring order when tossing three dice/coins and in this mis-interpretation came up with an answer in the case of three players which diverged from his own reverse recursive solution based on the principle of fair treatment at each node of his ragged tree.
Because Pascal's wrong-headed idea of Fermat's solution did not match his own, he jumped to the conclusion that what must be wrong in Fermat's method was the extension of the tree of possibilities beyond those parts which the game in hand required. Pascal consulted
Roberval on the likely legitimacy of this fully rolled out tree of possibilities and Roberval seems to have told Pascal that this is where Fermat is going wrong, namely that this 'false assumption' of theoretical play of zombie-games leads to bad results. It doesn't.
The evolution in time of a source of randomness was seen clearly by Fermat as separate from the rule, game or activity sitting on top of it. In this case the game was the 'first to get N wins' Modern derivatives when tree based methods are used all apply this same move. First the random process's set of possibilities are evolved on a lower, supporting layer, then the payoff of the contract is worked out at the terminal time horizon. Both in De Mere's game and with an option, there's a clearly defined termination point. With De Mere's game, the point happens when the first player reaches N wins. With options, the termination point is the expiry of the option. Gambler's ruin, as I'll discuss later, doesn't have such a straightforward termination point. So step 1 is to lay out all the possible states from now to the termination point, the tree of possibilities for the stochastic process. Then you work out the terminal value of the contract or game and use Pascal's fairness criterion to crawl back up the second tree, until you reach the 'now' point, which gives you the fair value of the contract. This is the essence of the finite difference solution set, and it works for path dependent and path independent pricings. The implications of the game is that the tree is re-combinant, which means the binomial coefficients become relevant when working out the probability that each path is traversed.
Fermat has a clearer and earlier conception of this separation. But Roberval and Pascal were right to flag this move up - what grounds did Fermat give for the move? In modern parlance, we can see that the stochastic process, often a stock price or a spot FX or a tradeable rate, is independently observable in the market. But back then, Pascal was struggling to separate the game from the source of randomness. F. N. David suggests that Pascal sets Roberval up as the disbeliever as a distancing mechanism for his own failure to grasp this point. Likewise, David suggests perhaps Pascal only solved his side of the problem after initial prompting from Fermat, in a letter which starts off the correspondence but which unfortunately no longer exists.
Of course, this isn't a solution of an unfinished game, but the fair value of the game at any point during its life. Each author I read seems clear in his mind that one other other of the great mathematicians' solution is preferred. Is this just ignorance, aesthetic preference masquerading as informed opinion? Yes, largely. But my own opinion is that the both solutions share many similarities - both need to evolve a tree of possibilities, a binary tree, for which the binomial coefficients come in handy as the number of steps increases. Both then involve evaluating the state of the game at the fixed and known horizon point. Fermat's tree is a set of possibilities of a stochastic process. His solution takes place exclusively at that final set of terminal nodes, but working out the ratio of the set of nodes in which player A is the winner over the total set of terminal nodes. Pascal's tree is the tree of game states. He reasons in a reverse iterative way until he reaches the start point, and the start point gives him his final answer. The arithmetic triangle could help both these men build their trees as the number of steps increases.
